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olga nikolaevna [1]
3 years ago
10

Help me What is 1/2 divided by 2/3

Mathematics
2 answers:
Anit [1.1K]3 years ago
6 0

Answer:

<h2>3/4</h2>

Step-by-step explanation:

\frac{1}{2}\div \frac{2}{3}\\\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}\\\\=\frac{1}{2}\times \frac{3}{2}\\\\\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}=\frac{1\times \:3}{2\times \:2}\\\\=\frac{3}{4}

ycow [4]3 years ago
3 0

Answer: 0.75 = 3/4

Step-by-step explanation: To solve, flip the second fraction and multiply the numerators and denominators together and simplify.

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You measure 22 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 5.
lara31 [8.8K]

Answer:

Step-by-step explanation:

We want to determine a 90% confidence interval for the true population mean textbook weight.

Number of sample, n = 22

Mean, u = 64 ounces

Standard deviation, s = 5.1 ounces

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

64 ± 1.645 × 5.1/√22

= 64 ± 1.645 × 1.087

= 64 ± 1.788

The lower end of the confidence interval is 64 - 1.788 = 62.21 ounces

The upper end of the confidence interval is 64 + 1.788 = 65.79 ounces

Therefore, with 90% confidence interval, the true population mean textbook weight is between 62.21 ounces and 65.79 ounces

3 0
3 years ago
Michael gets 52% better at press<br> ups. He used to be able to do 62.<br> How many can he do now?
maria [59]

Answer:

94 press ups

Step-by-step explanation:

62 x 1.52 = 94.24

               ≈94

4 0
3 years ago
When people make estimates, they are influenced by anchors to their estimates. A study was conducted in which students were aske
12345 [234]

Answer:

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

where μ1: mean calorie estimation for the cheesecake group and μ2: mean calorie estimation for the organic salad group.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first (P-value=0.0000002).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Suppose that the study was based on a sample of 20 people who thought about the cheesecake first and 20 people who thought about the organic fruit salad first, and the standard deviation of the number of calories in the cheeseburger was 128 for the people who thought about the cheesecake first and 140 for the people who thought about the organic fruit salad first.</em>

<em>At the 0.01 level of significance, is there evidence that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first?"</em>

<em />

This is a hypothesis test for the difference between populations means.

The claim is that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

The significance level is 0.01.

The sample 1 (cheese cake), of size n1=20 has a mean of 780 and a standard deviation of 128.

The sample 2 (organic salad), of size n2=20 has a mean of 1041 and a standard deviation of 140.

The difference between sample means is Md=-261.

M_d=M_1-M_2=780-1041=-261

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{128^2}{20}+\dfrac{140^2}{20}}\\\\\\s_{M_d}=\sqrt{819.2+980}=\sqrt{1799.2}=42.417

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-261-0}{42.417}=\dfrac{-261}{42.417}=-6.153

The degrees of freedom for this test are:

df=n_1+n_2-1=20+20-2=38

This test is a left-tailed test, with 38 degrees of freedom and t=-6.153, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.0000002) is smaller than the significance level (0.01), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

3 0
3 years ago
8 X 1 + 3 X (1/10) + 4 x (1/100)
Burka [1]
The answer I got was 417/50 hope this helps you
4 0
3 years ago
Read 2 more answers
Select the correct answer. Find the solution(s) for x in the equation below. x^2 -25 =0
Travka [436]

Answer:

Step-by-step explanation:

Find the roots of x^2-25=0 by solving for x.

Add 25 to both sides

x^2=25

Square root both sides

x=-5

x=5

4 0
2 years ago
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